Question

how do you integrate this?? ;[ http://puu.sh/7vLrj.jpg

Answer 1

whip out your trig theorems for integration

Answer 2

they don't work ;[

Answer 3

are you sure you can't use trig substitution?

Answer 4

I'm not sure.. have you worked this out already O_o

Answer 5

doesnt look pleasant
http://www.wolframalpha.com/input/?i=%5Cint+%5Csqrt%7B%5Ctan+x+%5Csec%5E2x%7D+dx

Answer 6

oh gosh

Answer 7

how do you do this without using that?

Answer 8

you pray that it goes away

Answer 9

LOL dont pray to hard aint gonna happen^^

Answer 10

hahah idk i never dealt wid these evil integrals before..
you're taught elliptic integrals in class is it ?

Answer 11

it's going to be on my final :,[

Answer 12

you're going to need lots of offerings to the gods that they may favor your fate

Answer 13

I usually pass out whenever I see these things

Answer 14

What math is this so i dont take it?

Answer 15

is it going to help you to use
tan^2 u + 1 = sec^2 u

Answer 16

it's integral calculus

Answer 17

lower division or upper? 1st year calc?

Answer 18

it's integration of trig f(x)

Answer 19

let me try by hand and see laughing out loud
wish me luck

Answer 20

do you happen to have the solution? im trying some crazy studd but it all seems legal

Answer 21

I don't have the answer ;[

Answer 22

u = tan x
du = sec^2 x dx

Answer 23

that doesn't work!

Answer 24

secx is what's present in the problem!

Answer 25

hey I am exploring...

Answer 26

xD

Answer 27

sorry :[

Answer 28

I am stressed out :c

Answer 29

ya
let u = tan x
du = sec^2 x

Answer 30

then?

Answer 31

\[\int\limits \sqrt{\tan x} \times \sqrt{\sec^2 x}\]

Answer 32

right?

Answer 33

welp nevermind...

Answer 34

how about multiplying it by 1?
\[\int\limits \sqrt{\tan x \sec^2 x} \times \frac{ \sqrt{1+\sec^2} }{ \sqrt{1+\sec^2} }\]

Answer 35

I think this is promising

Answer 36

u = 1+ sec^2
du = 2 tan x sec^2 x dx

Answer 37

\[\sqrt{2} \int\limits \sqrt{2\tan x \sec^2 x} dx\]

Answer 38

is this even legal? laughing out loud

Answer 39

actually it's supposed to be 4 as a constant

Answer 40

The first thing I thought of looking at this integral was a somewhat less frightening but similar looking problem with sqrt(tan x):
http://www.wolframalpha.com/input/?i=integral+of+sqrt%28tan+x%29+dx
http://math.ucsd.edu/~wgarner/math20b/int_sqrt_tan.htm
I have no other ideas for this at the time, though. :c

Answer 41

I tried that approach sqrt(tan x) earlier

Answer 42

without the square root in multiplication by 1

Answer 43

why not do it with substitution

Answer 44

don't you think that is what I am trying to do?

Answer 45

tanx = t
sec^2xdx =dt

Answer 46

that solution seems suspicious

Answer 47

so it becomes sqrt(t)dt

Answer 48

maybe from yahoo answer?

Answer 49

What?

Answer 50

I tried with
u = tan x
du = sec^2 dx
it did not get us anywhere because of the square root
you can't use a full substitution to get to du

Answer 51

look at my previous post, ankit

Answer 52

K hmm let me think then

Answer 53

keep googling for us, broth

Answer 54

brotha*

Answer 55

we are desperate :)

Answer 56

Dude i googled the differential for tanx as I don't remember it

Answer 57

>.<

Answer 58

@jim_thompson5910 @terenzreignz

Answer 59

let me pull out my notes
there might be some hope with long sub

Answer 60

why not first simplify tanx and sec^2x in terms of sin and cos

Answer 61

go ahead

Answer 62

everyone is free to shoot the problem with solutions

Answer 63

I have about a page of that trial already

Answer 64

Yeah just thinking out loud

Answer 65

weierstrass sub got me this...

t = tan (x/2)

sin x = (2t)/(1 + t²) cos x = (1 - t²)/(1 + t²)
tan x = 2t/(1 - t²) sec x = (1 + t²)/(1 - t²)
dx = 2 dt / (1 + t²)


integral sqrt[tan x sec² x] dx
integral sqrt[(2t)/(1 - t²)] (1 + t²)/(1 - t²) 2/(1 + t²) dt
integral sqrt[(2t)/(1 - t²)] 2/(1 - t²) dt
integral 2sqrt(2) sqrt(t)/(1 - t²)^(3/2) dt

somewhat interesting but still doesn't look too easy (infact reminds me a bit of the look of elliptic integrals). i'll come back to this after some sleep though. lol

Answer 66

^ this

Answer 67

show us the way, bat

Answer 68

this can, indeed, be integrated if we use tan(x/2)

Answer 69

the problem is yours

Answer 70

:D hell no

Answer 71

is that half-angle you're suggesting?

Answer 72

I am exploring secand-squared factor >.<

Answer 73

actually, forget what I said, I forgot the square root when I use tan(x/2) XD.

Answer 74

although the fraction turns out to be very nice looking, but can not be integrated :/

Answer 75

I know... I get the same thing earlier

Answer 76

this is suspicious
\[\int\limits \sqrt{\tan x \sec^2 x} \times \frac{ 1+\sec^2 }{ 1+sex^2 }\]
it's that pesky square root

Answer 77

not sex in the denominator i mean sec^2

Answer 78

ROFL. I make the same typo quite often myself

Answer 79

i'll think about this in my sleep
sorry @KNorne7592
i'll try again tomorrow

Answer 80

Not that this is likely to help...

Answer 81

The answer is 5

Answer 82

yeah it is 5

Answer 83

Hey really?

Answer 84

it's not 5

Answer 85

my bad it is 290

Answer 86

that's not the answer

Answer 87

ok ask me another question

Answer 88

if you're not going to help, go away

Answer 89

Do you have a rough estimate of what integral tricks you've learned so far (like u-sub up to things like half-angle tangent sub and such)? Just curious to know roughly what to expect is fair game.

Answer 90

none of them are working in this problem

Answer 91

What have you learnt so far, like what chapter are you on?